Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
A. 1.13 s; 69.75 ft


B.1.13 s; 29.25 ft


C.2.25 s; 9 ft


D.1.13 s; 31.5 ft

Answer 1

Answer:  B) 1.13 s; 29.25 ft

Step-by-step explanation:

Height of the ball h in feet after t seconds is given by the function

$h=-16t^{2}+36t+9$

Now at maximum height $\frac{\partial h}{\partial t}=0=>\frac{\partial( -16t^{2}+36t+9)}{\partial t}=0$

=>$\therefore -32t+36=0=>t=1.125\, seconds$

After rounding to nearest hundredth,$t\simeq 1.13\, seconds$

Height , $h_{max}=-16t^{2}+36t+9$

$=>h_{max}=[-16(1.125)^{2}+36\times 1.125+9] feet=29.25 feet$

Therefore , t=1.13 s and  $h_{max}=29.25 feet$

Thus correct option is B) 1.13 s; 29.25 ft

Answer 2

B is ur answer hope i helped